Solve/word problem?

Question

tickets to the bryce semmesterin concert at the assembly hall cost $18 for A section, $8 for B section and $6 for C section. All 16,000 tickets were sold and total receipts were $112,000. There are twice as many seats in B section as in A section. How many seats are there in section C?

Answer 1

18A + 8B + 6C = 112000 ==> eq. 1
A + B + C = 16000 ==> eq. 2
B = 2A ==> eq. 3

eq. 3 into eq. 1 and eq. 2:

18A + 8(2A) + 6C = 112000 ==> 34A + 6C = 112000
A + 2A + C = 16000 ==> 3A + C = 16000

Multiply eq. 5 by -6:

-18A - 6C = -96000

Add eq. 4 and eq. 6:

16A = 16000
A = 1000

C = 16000 - 3(1000)
C = 13000

B = 2(1000)
B = 2000

Answer 2

let a be the number of seats in section A, b in section B and c in section C

a + b + c = 16000

18 dollars for a
8 dollars fo b
6 dollars fo c

18a + 8b + 6c = 112,000

There are twice as many seats in B section as in A section tells you: b = 2a

Now you have 3 equations:

a + b + c = 16,000
18a + 8b + 6c = 112,000
b = 2a


get rid of c in the first two equations
-6(a + b + c = 16,000)
18a + 8b + 6c = 112,000

-6a - 6b - 6c = -96,000
18a + 8b + 6c = 112,000
------------------------------------
12a + 2b = 16,000

we know that b = 2a
12a + 2(2a) = 16,000
12a + 4a = 16,000
16a = 16000
a = 1000

b = 2a
b = 2(1000)
b = 2000

c = 16000 - (2000 + 1000)
c = 13,000

There are 13,000 seats in section C

Answer 3

Let A,B,C be the respective number of seats in each section
18A + 8B + 6C = 112000
A + B + C = 16000
B = 2A
Now just put in the value of B and solve for C

Answer 4

B = 2A
A + B + C = 16000
or 3A + C = 16000
or C = 1600O - 3A (1)

18A + 8B + 6C = 112000
or 3C = 56000 - 15A (2)

Now solve (1) and (2) for C.
C = 3000

Hope this helps.

[e-mail]

Answer 5

A ---$18--- x seats
B----$8-----2x seats
C----$6------y seats

18x + 16 x + 6y = 112000

3x + y = 16000

34x + 6y = 112000
-18x - 6y = - 96000----ADD
16x = 16,000
x = 1,000
y = 13,000

13,000 seats in section C